RUSSEL | Usability Support System | Video Utils library
kandi X-RAY | RUSSEL Summary
kandi X-RAY | RUSSEL Summary
## Table of Contents.
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Top functions reviewed by kandi - BETA
- Display the Renders
- Hook the status of the zip file and extract the contents of the zip file
- Process the pending uploads
- Build an empty tile for an empty file
- Display the Utility screen
- Fills the group details
- Creates a group
- Delete a group
- Creates filter based on asset filter
- Gets the ISD activity packet
- Initializes the node
- Display the login screen
- Login with the given username and password
- Creates a zip package
- Updates the missing file information
- Parses ESBP packet
- Process the serverZipIds from the server side stream
- Rounds a number to the login page
- Setup the template hooks
- Displays the Renders
- Display the Feature home screen
- Renders the EPS Edit Screen
- Displays the RRS menu
- Assigns the listeners to the object click
- Display the results
- Display the Utility screen
RUSSEL Key Features
RUSSEL Examples and Code Snippets
Community Discussions
Trending Discussions on RUSSEL
QUESTION
According to Artificial Intelligence A Modern Approach - Stuart J. Russell , Peter Norvig (Version 4), space complexity of BFS is O(b^d), where 'b' is branching factor and 'd' is depth.
Complexity of BFS is obtained by this assumption: we store all nodes till we arrive to target node, in other word: 1 + b + b^2 + b^3 + ... + b^d => O(b^d)
But why should we store all nodes? don't we use queue for implementation?
If we use queue, don't need to store all nodes, because we enqueue and dequeue some nodes in steps, then when we find target node(s), we can say some nodes are in queue (but not all of them).
Is my understanding wrong?
...ANSWER
Answered 2022-Apr-10 at 06:16At any moment while we apply BFS, the queue would have at most two levels of nodes, for example if we just started searching in depth d, then the queue now contains all nodes at depth d and as we proceed the queue would finish all nodes at depth d and have all nodes at depth d+1, so at any moment we have O(b^d) space.
Also 1+b+b^2+...+b^d = (b^(d+1)-1)/(b-1).
QUESTION
I'm new to bash shell and I have to do a script with a csv file.
The file is a list of the participants, countries, sports and medals achieved.
when executing the script, I should give as parameters the nationality
(column 3) and the sport
(column 8). The script should return the amount of participants of that country for that sport, and the amount of medals achieved.
The amount of medals achieved is the sum of the columns "gold" "silver" "bronze" of each row which are columns 9,10 and 11.
I cannot use grep, awk, sed or csvkit.
So far, I have this code but I'm stuck with the medal counting part.
...ANSWER
Answered 2022-Apr-02 at 19:21Here is a pure bash implementation. Build a hash from field name to position ($h
):
QUESTION
standings
...ANSWER
Answered 2022-Mar-29 at 15:45You can't do assignment (A=B) in a comprehension. You could possibly use the update function. For example:
QUESTION
I can't seem to fix this error which is only happening after I've deployed and it's driving me crazy. I've tried everything from all other stack overflow suggestions with no luck. I get no errors on local, however when I deploy to Heroku I get the error "Error: "line" is not a registered controller." Any ideas?
...ANSWER
Answered 2022-Mar-23 at 06:34You need to import and register the LineController
QUESTION
I'm trying to create custom find method using Array.prototype
, I tried a lot of way but i just couldn't break the loop for the first match:
ANSWER
Answered 2022-Jan-27 at 16:54Well right now you just have a for loop, if you want to find any element that satisfies a condition, you have to return a boolean from your callback and return the element, for which that returned boolean is true
. If you want the index of the element you found and the initial array to be returned as well, you can return an object and destructure it:
QUESTION
I'm using Spring Integration using DSL for handling communication between JMS and REST service. The requirement is that messages should be redelivered indefinetly. In one case, I have to sequentially execute two operations. If first one fails, the second one shouldn't be executed, but in case it's 4xx error I shouldn't try to redeliver it. My code looks like this:
...ANSWER
Answered 2022-Jan-19 at 20:42That is the default behavior; the second subscriber won't be called unless the ignoreFailures
property is true
(it is false
by default).
You need to show the upstream flow, but to "catch" the exception you need to add an error channel to the (presumably) message-driven inbound adapter and handle the exception there.
QUESTION
I have 2 symbols in my backtest. I am adding indicator donchianchannel. However when i plot it, add_TA( mktdata$high.DCH, col = 6, lwd = 1.5,on=TRUE) does not pass the associated symbol's data and i end up getting the same data plotted for both the symbols.
...ANSWER
Answered 2022-Jan-12 at 02:00Try this instead, at the end of your example :
QUESTION
Im working through some self-join examples and I am drawing a blank on the following example. Its the last example at the following link Self-Join Example
...ANSWER
Answered 2021-Nov-26 at 15:51If you didn't have any condition on employee ID at all you'd end up with records where a self-match had occurred, e.g. the results would show "Gracie Gardner was hired on the same day as Gracie Gardner"
We could then put ON e1.employee_id <> e2.employee_id
- this would prevent Gracie matching with Gracie, but you'd then find "Gracie Gardner was hired on the same day as Summer Payne" and "Summer Payne was hired on the same day as Gracie Gardner" - i.e. you'd get "duplicate records" in terms of "person paired with person", each name being mentioned both ways round
Using greater than prevents this, and effectively means that any given pair of names only appears once. Because Gracie's ID is less than Summer's, you'll get Gracie in e1
paired with Summer in e2
but you won't get Summer in e1
paired with Gracie in e2
Another way of visualizing it is with a square/matrix
QUESTION
I am trying on reactor-kafka for consuming messages. Everything else work fine, but I want to add a retry(2) for failing messages. spring-kafka already retries failed record 3 times by default, I want to achieve the same using reactor-kafka.
I am using spring-kafka as a wrapper for reactive-kafka. Below is my consumer template:
...ANSWER
Answered 2021-Oct-01 at 09:40Previously while retrying I was using the below approach:
QUESTION
I have few images like these,
I can extract the names and roles from these images using a ocr tool like tesseract from python, but I want to preserve the hierarchy along the way.
Please provide some interesting way to solve this problem. I am not able to think of one proper approach to the problem.
...ANSWER
Answered 2021-Sep-15 at 21:28Visualization of results:
Approach:
- box borders and connecting lines ("nets") have certain color/brightness
- work with masks, lists of contours, label maps
- calculate overlap/intersection
- at overlap, check what box and what net participate
Written for the one specific image you provided. The other one is too low-resolution.
For less favorable input data, this would need adapting. That shouldn't be difficult though. Just different thresholds and whatnot.
This should also already work with non-box nodes (e.g. circles/ellipses).
You can figure out the OCR part. This approach and code gives you the individual boxes that you can pass to OCR.
Output:
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install RUSSEL
You can use RUSSEL like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the RUSSEL component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .
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